A ball of mass 070 kg is attached to a vertical post by strings of length L_(1)=1.1m and L_(2)=1.7 m, as shown in (Figure 1). The ball is set moving in a horizontal circle. Part B Part A Determine the maximum speed for the lower string to be slack (tension in cord=0) Express your answer with the appropriate units. v=5.05 (m)/(s) Incorrect; Try Again; 5 attempts remaining

To determine the maximum speed for the lower string to be slack, we need to consider the forces acting on the ball when the tension in the lower string becomes zero. At this point, the only forces acting on the ball are gravity and the tension in the upper string.<br /><br />The centripetal force required to keep the ball moving in a circular path is provided by the vertical component of the tension in the upper string. When the lower string is slack, the vertical component of the tension in the upper string must balance the weight of the ball.<br /><br />Let's denote the mass of the ball as \( m \), the length of the upper string as \( L_1 \), and the length of the lower string as \( L_2 \). The vertical component of the tension in the upper string is given by \( T \sin(\theta) \), where \( \theta \) is the angle between the upper string and the vertical.<br /><br />The weight of the ball is given by \( mg \), where \( g \) is the acceleration due to gravity.<br /><br />Setting the vertical component of the tension equal to the weight of the ball, we have:<br /><br />\( T \sin(\theta) = mg \)<br /><br />Since the tension in the upper string provides the centripetal force, we can also write:<br /><br />\( T = \frac{mv^2}{L_1} \)<br /><br />Substituting the expression for \( T \) into the first equation, we get:<br /><br />\( \frac{mv^2}{L_1} \sin(\theta) = mg \)<br /><br />Simplifying, we have:<br /><br />\( \frac{v^2}{L_1} \sin(\theta) = g \)<br /><br />Solving for \( v \), we get:<br /><br />\( v = \sqrt{g L_1 \sin(\theta)} \)<br /><br />To find the maximum speed, we need to maximize \( \sin(\theta) \), which occurs when the angle \( \theta \) is 90 degrees (i.e., when the ball is at the lowest point of its motion).<br /><br />Therefore, the maximum speed is given by:<br /><br />\( v = \sqrt{g L_1} \)<br /><br />Substituting the given values, we have:<br /><br />\( v = \sqrt{9.8 \, \text{m/s}^2 \times 1.1 \, \text{m}} \)<br /><br />\( v = \sqrt{10.78 \, \text{m}^2/\text{s}^2} \)<br /><br />\( v = 3.28 \, \text{m/s} \)<br /><br />Therefore, the maximum speed for the lower string to be slack is approximately \( 3.28 \, \text{m/s} \).